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Homework Help: Prove irrational

  1. Mar 13, 2010 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    I need to prove that [tex]\sqrt{3}[/tex] is irrational.


    3. The attempt at a solution
    The prior problem was to show [tex]\sqrt{2}[/tex] is irrational and the solution had to do with a contradiction that each number must be even or something (frankly, I didn't understand it too well). But I don't see how I can apply the same idea to this one, since evenness is not applicable here.

    All i can really do so far is assume [tex]\sqrt{3}=\frac{p}{q}[/tex] where p and q are integers.
    Now, by squaring, [tex]3q^2=p^2[/tex].

    I am completely lost here though...

    Some help please?
     
  2. jcsd
  3. Mar 13, 2010 #2

    CompuChip

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    The proof of case with 2 depends on the fact that either p is divisible by 2 (and then you can show that q is too, which contradicts the assumption that you choose them to be "minimal"), or it is not (and since [itex]2q^2 = p^2[/itex] tells you that it is, you have a contradiction).

    The proof for 3 is similar: p is either divisible by 3, and then so is q contradicting our assumption; or p is not divisible by 3 and then it must be divisible by 3 which is also impossible.

    Perhaps it is a good idea to go through the proof of the irrationality of [itex]\sqrt{2}[/itex] once more, and try to find out what part confuses you. Can you try and explain it to us?
     
  4. Mar 13, 2010 #3

    Mentallic

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    With the [tex]\sqrt{2}[/tex] question the solution says that if [tex]2q^2=p^2[/tex] then [tex]p^2[/tex] is even, thus p is even, so we can put p=2r for integer r, so now we have [tex]2q^2=4r^2[/tex] which also implies that q is even but this is the contradiction.

    Actually.. while writing this I tested to see what the difference is for rational cases like [tex]\sqrt{4}[/tex] and [tex]\sqrt{9}[/tex]. Yes, now I understand it. It wasn't the evenness of [tex]\sqrt{2}[/tex]'s example that was special, it was the fact that it was divisible by 2 (the same thing I know :smile:).
     
  5. Mar 13, 2010 #4

    CompuChip

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    Yes, in this case it is the same.
    But in the general case [itex]\sqrt{n}[/itex], you want to say something about the divisibility by [itex]n[/itex].

    (Note that in passing you have also shown that the square of an even number is always divisible by not just 2, but by 4).
     
  6. Mar 13, 2010 #5

    If you add an additional information that gcd(p,q)=1 , which also means the greatest common divisor of p and q is 1 . However for [tex]\sqrt{2}[/tex] you can draw the conclusion that the gcd(p,q) is atleast 2 and for [tex]\sqrt{3}[/tex] gcd(p,q) is atleast 3. In fact you can prove a generalised one analogously which states the [tex]\sqrt{prime}[/tex] will lead you to gcd(p,q) is atleast the prime number , yet definition of prime states it cannot be 1 , will leads to the irrationality of all prime.
     
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